Question

A spring of spring constant $5 \times 10^3 Nm^{-1}$ is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is

• A. 12.50 N-m
• B. 18.75 N-m
• C. 25.00 N-m
• D. 6.25 N-m

Answer 1

Answer: (B)

Work done $W_1=\frac{1}{2}k \times x^2_1$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{1}{2}\times5\times10^3 \times(5\times10^{-2})^2$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 6.25 J$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, W_2=\frac{1}{2}k(x_1+x_2)^2$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{1}{2} \times5\times10^3 (5\times10^{-2}+5\times 10^{-2})^2$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =25 J$
Net work done $\, \, \, \, \, \, \, \, \, \, \, = W_2 - W_1$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =25-6.25$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 18.75 J =18.75 N-m$